What is the integral of (2t)/(1-t^2) ?

The integral of (2t)/(1-t^2) is -ln|1-t^2|+C

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integral of (2t)/(1-t^2)

\int \frac{2 t}{1 - t ^{2}} d t

- ln 1 - t^{2} + C

Solution steps

\int \frac{2 t}{1 - t ^{2}} d t

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{t}{1 - t ^{2}} d t

Apply u-substitution

: \int - \frac{1}{2 u} d u

= 2 \cdot \int - \frac{1}{2 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= 2 (- \frac{1}{2} ln ∣ u ∣)

Substitute back

u = 1 - t^{2}

= 2 (- \frac{1}{2} ln 1 - t^{2})

Simplify

2 (- \frac{1}{2} ln 1 - t^{2}) : - ln 1 - t^{2}

= - ln 1 - t^{2}

Add a constant to the solution

= - ln 1 - t^{2} + C

n = 1 \sum \infty n \cdot 2^{- n}

- 5 t^{2} y^{^{''}} - 2 t y^{^{'}} + 14 y = 0

f (x) = π x

x^{2} e^{x} - 2 x e^{x} + 5 e^{x}

x \to 0 lim (\frac{x ^{2} + 5 x}{x})

What is the integral of (2t)/(1-t^2) ?
The integral of (2t)/(1-t^2) is -ln|1-t^2|+C